Problem: Let $f$ be defined by  \[f(x) = \left\{
\begin{array}{cl}
2-x & \text{ if } x \leq 1, \\
2x-x^2 & \text{ if } x>1.
\end{array}
\right.\]Calculate $f^{-1}(-3)+f^{-1}(0)+f^{-1}(3)$.
Explanation: The number $f^{-1}(-3)$ is the value of $x$ such that $f(x) = -3$.  Since the function $f$ is defined piecewise, to find this value, we must consider both cases $x \le 1$ and $x > 1$.

If $x \le 1$ and $f(x) = -3$, then $2 - x = -3$, which leads to $x = 5$.  But this value does not satisfy the condition $x \le 1$.  If $x > 1$ and $f(x) = -3$, then $2x - x^2 = -3$, or $x^2 - 2x - 3 = 0$.  This equation factors as $(x - 3)(x + 1) = 0$, so $x = 3$ or $x = -1$.  The only value that satisfies the condition $x > 1$ is $x = 3$, so $f^{-1}(-3) = 3$.

Next, we compute $f^{-1}(0)$, which is the value of $x$ such that $f(x) = 0$.

If $x \le 1$ and $f(x) = 0$, then $2 - x = 0$, which leads to $x = 2$.  But this value does not satisfy the condition $x \le 1$.  If $x > 1$ and $f(x) = 0$, then $2x - x^2 = 0$, or $x^2 - 2x = 0$.  This equation factors as $x(x - 2) = 0$, so $x = 0$ or $x = 2$.  The only value that satisfies $x > 1$ is $x = 2$, so $f^{-1}(0) = 2$.

Finally, we compute $f^{-1}(3)$, which is the value of $x$ such that $f(x) = 3$.

If $x \le 1$ and $f(x) = 3$, then $2 - x = 3$, which leads to $x = -1$.  Note that this value satisfies the condition $x \le 1$.  If $x > 1$ and $f(x) = 3$, then $2x - x^2 = 3$, or $x^2 - 2x + 3 = 0$.  This equation can be written as $(x - 1)^2 + 2 = 0$, which clearly has no solutions, so $f^{-1}(3) = -1$.

Therefore, $f^{-1}(-3) + f^{-1}(0) + f^{-1}(3) = 3 + 2 + (-1) = \boxed{4}$.

[asy]
unitsize(3mm);
defaultpen(linewidth(.7pt)+fontsize(8pt));
import graph;

draw((-8,0)--(8,0),Arrows(4));
draw((0,-8)--(0,8),Arrows(4));

real f(real x) {return 2-x;}
real g(real x) {return 2x-x^2;}

real x;

draw(graph(f,-5,1),BeginArrow(4));
draw(graph(g,1,4),EndArrow(4));

real eps = 0.2;

draw((-eps,3)--(eps,3));
draw((-eps,0)--(eps,0));
draw((-eps,-3)--(eps,-3));

dot("$(-1,3)$",(-1,3),SW);
dot("$(2,0)$",(2,0),NE);
dot("$(3,-3)$",(3,-3),E);

label("$f(x)$",(1.5,8.5));
label("$x$",(8.5,-1));
[/asy]